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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 15.53

Calculate the pH of 0.105 M CCl3COOH (trichloroacetic acid, pKa = 0.52).




Answer:

Write the reaction, write the mass action expression and find Ka, set up a table of concentrations, plug the equilibrium concentrations into the mass action expression, and then solve.

CCl3COOH(aq)
+
H2O(l)
H3O+(aq)
+
CCl3COO (aq)
Initial
0.105
0
0
Change
– x
+ x
+ x
Equilibrium
0.105 – x
x
x

[HA]init/Ka = 0.105/0.30 = 0.35 < 100 so no approximation

Using the quadratic equation gives

x = –0.38 or x = 0.082; only the positive value makes sense so

x = [H3O+] = 0.082 M

pH = –log[H3O+] = –log[0.082] = 1.09